1. (教材练习变式)在运用乘法公式计算$(2x - y + 3)(2x + y - 3)$时,下列变形正确的是()

A.$[(2x - y) + 3][(2x + y) - 3]$
B.$[(2x - y) + 3][(2x - y) - 3]$
C.$[2x - (y + 3)][2x + (y - 3)]$
D.$[2x - (y - 3)][2x + (y - 3)]$

1. D

2. 下列各式中,能用完全平方公式计算的是()

A.$(2a - 4b)(-2a - 4b)$
B.$(a + 4b)(a + 4b)$
C.$(a - 4b)(a + 4b)$
D.$(2a - 4b)(4a + 2b)$

2. B

3. 下列计算正确的是()

A.$(x - y)(-y - x) = y^{2} - x^{2}$
B.$(2x - y)(y - 2x) = -y^{2} - 4x^{2}$
C.$(2a - 1)^{2} = 4a^{2} - 2a + 1$
D.$(3 - x)^{2} = 9 - x^{2}$

3. A 解析:$(x-y)(-y-x)=(-y)^{2}-x^{2}=$
$y^{2}-x^{2}$,故 A 选项正确;$(2x-y)(y-2x)=-4x^{2}+4xy-y^{2},$
故 B 选项错误;$(2a-1)^{2}=4a^{2}-4a+1$,故 C 选项错误;(3-
$x)^{2}=9-6x+x^{2}$,故 D 选项错误.

4. 填空:(1)$(x - \frac{1}{2}y) ·\_\_\_\_\_\_= x^{2} - \frac{1}{4}y^{2}$; (2)$m^{2} - 4m +\_\_\_\_\_\_= (m -\_\_\_\_\_\_)^{2}$.

4. (1)$(x+\frac {1}{2}y)$ (2)4
2

5. 若$a^{2} + b^{2} = 8$,$ab = 2$,则$(a - b)^{2} =$.

5. 4 解析:$(a-b)^{2}=a^{2}+b^{2}-2ab=8-2×2=4$.

$(a - b)^2 = a^2 + b^2 - 2ab = 8 - 2×2 = 4$

6. 如果$(2a + 2b + 1)(2a + 2b - 1) = 63$,那么$a + b$的值为.

6. ±4
解析:$\because (2a+2b+1)(2a+2b-1)=63,\therefore (2a+2b)^{2}-1^{2}=$
$63,\therefore (2a+2b)^{2}=64,\therefore 2a+2b=\pm 8$,两边同时除以 2,得$a+$
$b=\pm 4$.

7. 计算:
(1)$(2x + 3)(4x^{2} + 9)(2x - 3)$;
(2)$(x + 1)(x - 1)(x^{2} + 1)(x^{4} + 1)$;
(3)$(x + 2y - 3)(x - 2y + 3)$;
(4)$(m + 2n - 3)(m - 2n - 3)$;
(5)$(m - 2n)^{2}(-m - 2n)^{2}$;
(6)$(-a + 1)(a + 1)(a^{2} - 1)$.

7. (1)原式$=(2x+3)(2x-3)(4x^{2}+9)=(4x^{2}-$
$9)(4x^{2}+9)=16x^{4}-81$. (2)原式$=(x^{2}-1)(x^{2}+1)(x^{4}+$
$1)=(x^{4}-1)(x^{4}+1)=x^{8}-1$. (3)原式$=[x+(2y-3)][x-$
$(2y-3)]=x^{2}-(2y-3)^{2}=x^{2}-4y^{2}+12y-9$. (4)原式=
$[(m-3)+2n][(m-3)-2n]=(m-3)^{2}-(2n)^{2}=m^{2}-6m+$
$9-4n^{2}$. (5)原式$=[(m-2n)(-m-2n)]^{2}=[(-2n)^{2}-$
$m^{2}]^{2}=(4n^{2}-m^{2})^{2}=16n^{4}-8m^{2}n^{2}+m^{4}$. (6)原式$=-(a-$
$1)(a+1)(a^{2}-1)=-(a^{2}-1)^{2}=-(a^{4}-2a^{2}+1)=-a^{4}+$
$2a^{2}-1$.

8. 先化简,再求值:(1)$(a + 2)^{2} + (a + 1)(a - 1) - a(2a - 1)$,其中$a = -\frac{4}{5}$.
(2)$(x - y)^{2} - (2x + y)(2x - y) + 3x(x + y)$,其中$\vert x + 3\vert + (y - 2)^{2} = 0$.

8. (1)原式$=a^{2}+4a+4+a^{2}-1-2a^{2}+a=5a+3.$
当$a=-\frac {4}{5}$时,原式$=5×(-\frac {4}{5})+3=-1$. (2)原式=
$x^{2}-2xy+y^{2}-(4x^{2}-y^{2})+3x^{2}+3xy=x^{2}-2xy+y^{2}-4x^{2}+$
$y^{2}+3x^{2}+3xy=xy+2y^{2}.\because |x+3|+(y-2)^{2}=0,\therefore x+3=0,$
$y-2=0,\therefore x=-3,y=2$.
∴原式$=-3×2+2×2^{2}=2.$